Intermediate Algebra, Chapter 2, 2.7, Section 2.7, Problem 32

Solve the inequality $|5 - x| > 3$, and graph the solution set.

The absolute value inequality is rewritten as

$5-x > 3$ or $5-x < -3$,

because $5 - x$ must represent a number that is more than $3$ units from on either side of the number line. We can solve the compound inequality.


$\begin{equation} \begin{aligned} |5 - x| > & 3 && \qquad \text{or} &&& 5 - x < & -3 && \\ -x > & -2 && \qquad \text{or} &&& -x < & -8 && \text{Subtract } 5 \\ x < & 2 && \qquad \text{or} &&& x > & 8 && \text{Divide $-1$. Reverse the inequality symbols.} \end{aligned} \end{equation}$


The solution set is $(- \infty, 2) \cup (8, \infty)$.